how to get the record which is at 70-75 position

[Former Member]

hi guys,

i have a record as follows.

'5200ALLSTATE INSURANDIRECT DEPOSIT 2360719665CCDPAYABLES 080307 1053000210000001'.

as it is starting with '5' i should write the coding as if the record starts with '5' then i should pick the date which is at 70-75 position.can anybody provide me logic or coding for this.

for suitable answers reward points will b there.

very urgent.

[Former Member]

Are you sure is always going to be at 70-75 positions?

That 'ALLSTATE INSURANDIRECT DEPOSIT ' looks like a variable string

If record(1) = '5'.
date = record(6)+69.
endif.

[Former Member]

hi ramiro,

thanx for ur reply first.

yes,

it is always at 70-750 position.

i will try this code.

[Former Member]

romario,

the record which i told is in a file in 2nd row.so,this logic is not working for this.after reading ist row,it is coming out.

can u give the solution for this.

[Former Member]

But is going to remain in the 70-75 position?

How are you reading the file? Are you using gui_download?

Can you post the code?

The solution remains the same, you only add a loop to the file table.

Loop at it_arch.
if it_arch(1) = '5'.
date = it_arch+69(6).
endif.
endloop.

[Former Member]

Not exactly the same. It was:

If record(1) = '5'.
date = record(6)+69.
endif.

Your new code looks correct, but I waould avoid offsets. Future generations maintaining it would thank you.

Rob

[Former Member]

Hello,

You could try something like this:

if l_yourrecord(1) = '5'.

l_date = l_yourrecord+69(6) .

endif.

Regards

Greg Kern

[Former Member]

Hi, You can try this.

data: firts type c,

date(5) type c.

firts = <record>-field+0(1).

if ( first = '5' ).

date = <record>-field+69(5).

endif.

regards.

[Former Member]

Have a look at this:

DATA f1(750) VALUE
'5200ALLSTATE INSURANDIRECT DEPOSIT 2360719665CCDPAYABLES 080307 1053000210000001'.

DATA: BEGIN OF f2,
        a(1),
        b(69),
        date(6),
        c(674),
      END   OF f2.

MOVE f1 TO f2.

IF f2-a = '5'.
  WRITE: /001 f2-date.
ENDIF.

Rob

Edited by: Rob Burbank on Mar 11, 2008 4:14 PM

[Former Member]

hello

i have a question the date field suppose to be 8 in length but your requiremnt is only 6 in lenth anyways to acheive the requirement you can follow the below code.

code..

DATA: T_tab TYPE MATCH_RESULT.

data:v_date type sy-datum.

Find first occurences of '5' in v_variable results t_tab.

if t_tab-offset EQ '1'.

v_date = v_variable+69(6).

endif.

[Former Member]

srini

You will note in his data string, the date is in MMDDYY format and is likely from an outside source as from a bank. It may be necessary for him to compute the century to complete the YYYYMMDD format you are referring to.